Motion in a plane NCERT Solutions Class 11 Physics - Solved Exercise Question 4.16

Question 4.16:

A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?


Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R = u2 Sin 2θ / g
100 = u2 Sin 900 / g
u2 / g = 100   ....(i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = –g
Using the third equation of motion:
v2 - u2 = -2gH
H = u2 / 2g  =  100 / 2  =  50 m

No comments:

Post a Comment